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AS math example - Thanks

  1. Jul 26, 2005 #1
    Find the numbers M and N such that:

    5+4x -x^2 = m - (x - n)^2
    for all real values of x.

    If someone could work this out, Then ill show my workings and ask questions (its to do with the arrangements of N^2 after expanding brackets mainly.

    Thanks in advanced,
  2. jcsd
  3. Jul 26, 2005 #2


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    Do you know how to complete the square?

    ax² + bx + c
    = a[x² + (b/a)x] + c
    = a[x² + (b/a)x + (b/2a)² - (b/2a)²] + c
    = a[x² + (b/a)x + (b/2a)²] - a(b/2a)² + c
    = a[x + (b/2a)]² + (c - b²/4a)

    Complete the square for the left side of your equation, and it will look like the right side, but instead of m and n, you'll have numbers, and those are obviously the numbers you're trying to find.
  4. Jul 26, 2005 #3
    Complete the square, to put the equation in the form, (x - n)^2 + m , and then you have your values for n and m.

    multiply both sides by -1 , to get, x^2-4x-5 which can be written as (x-2)^2-9 , which is in the form (x - n)^2 + m

    so n=2, m=-9
  5. Jul 26, 2005 #4


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    Recheck that, roger. The problem states: 5+4x -x^2 = m - (x - n)^2, not 5+4x -x^2 = m + (x - n)^2
    Viet Dao,
  6. Jul 26, 2005 #5

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